Differential Equation - Acorn Prob

[InterserveVB]

I have no clue how to do this. Any ideas?

An acorn falls into a pond, creating a circular ripple whose area is increasing at a constant rate of 5pi m^2 / sec. When the radius of the circle is 4 m, at what rate is the diameter of the circle changing?

 

[Daniel_Feldman]

a=pi(r^2)

r=d/2

a=pi(d^2/4)=(pi/4)d^2

Using related rates

da/dt=2(pi/4)d(dd/dt)=(pi/2)d(dd/dt)

da/dt=5pi

r=4

d=2r=8

5pi=(pi/2)(8)(dd/dt)

5pi=(4pi)(dd/dt)

dd/dt=5/4 m/sec